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76.minimum-window-substring.cpp
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82 lines (75 loc) · 2.15 KB
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// Tag: Hash Table, String, Sliding Window
// Time: O(M + N)
// Space: O(N)
// Ref: -
// Note: -
// Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string "".
// The testcases will be generated such that the answer is unique.
//
// Example 1:
//
// Input: s = "ADOBECODEBANC", t = "ABC"
// Output: "BANC"
// Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.
//
// Example 2:
//
// Input: s = "a", t = "a"
// Output: "a"
// Explanation: The entire string s is the minimum window.
//
// Example 3:
//
// Input: s = "a", t = "aa"
// Output: ""
// Explanation: Both 'a's from t must be included in the window.
// Since the largest window of s only has one 'a', return empty string.
//
//
// Constraints:
//
// m == s.length
// n == t.length
// 1 <= m, n <= 105
// s and t consist of uppercase and lowercase English letters.
//
//
// Follow up: Could you find an algorithm that runs in O(m + n) time?
//
class Solution {
public:
string minWindow(string s, string t) {
int n = s.size();
int k = t.size();
unordered_map<char, int> counter;
for (auto c : t) {
counter[c]++;
}
int i = 0;
int count = 0;
int l = -1;
int r = -1;
for (int j = 0; j < n; j++) {
if (counter.count(s[j]) > 0) {
if (counter[s[j]] > 0) {
count += 1;
}
counter[s[j]] -= 1;
}
while (count == k) {
if (l == -1 || j - i < r - l) {
l = i;
r = j;
}
if (counter.count(s[i]) > 0) {
counter[s[i]] += 1;
if (counter[s[i]] > 0) {
count -= 1;
}
}
i += 1;
}
}
return l >= 0 ? s.substr(l, r - l + 1): "";
}
};